Empirical and Molecular Formulas

Question

When 3.145 grams of a hydrocarbon, C_xH_y, were burned in a combustion analysis apparatus, 9.207 grams of CO₂ and 5.655 grams of H₂O were produced.

In a separate experiment, the molar mass of the compound was found to be 30.07 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

empirical:

molecular:

Answer 1

Find moles of carbon (C):

9.207 g CO2 x 1 mol CO2/44 g x 1 mol C / mol CO2 = 0.209 moles C

Find moles of hydrogen (H):

5.655 g H2O x 1 mol H2O/18 g x 2 mol H / mole H2O = 0.628 moles H

To get whole numbers, divide by 0.209 and arrive at...

moles C = 0.209/0.209 = 1.00

moles H = 0.628/0.209 = 3.00

Empirical formula = CH₃

To find the molecular formula (some multiple of the empirical formula) we see how many of the empirical formulas it takes to make a molecular formula. The molar mass of the empirical formula is 15 g/mol

Molar mass of molecular formula is 30.07, so it takes TWO empirical to make 1 molecular.

Molecular formula = C₂H₆